Relativistic Twins

[Batalov]

(That was chapter Two of the Big Belly polygraph)
Moar! Moar! ...please don't stop writing.

[ewmayer]

I think a monoid is what they call an adenoid before its marriage. Just like a moron who loses a lot of weight becomes a lesson.

Ha, ha, "total RT displacement = 0" ... still chuckling over that one.

Hey, Dave, any word back from Physical review Letters on your revolutionary moneygram on cosmetology yet?

[davar55]

Originally Posted by ewmayer

I think a monoid is what they call an adenoid before its marriage. Just like a moron who loses a lot of weight becomes a lesson.

Ha, ha, "total RT displacement = 0" ... still chuckling over that one.

Hey, Dave, any word back from Physical review Letters on your revolutionary moneygram on cosmetology yet?

(3) No, not yet. They must be busy ...

(2) Glad you've got the chuckles. But how is that statement wrong?
Inquiring mind needs to know.

(1) Not before his marriage, but after his divorce (more or less).

Thanks again for your so insightful criticism.

(0) I can certainly understand the desire to cling to one's
preconceived notions, even in the face of overwhelming
evidence/argumentation. :)

[ewmayer]

"If I take a long drive in my car and end up at the same place I started, my odometer should show no change, right?"

Dave, any instinct I might have had to "be patient with the physics neophyte" does not apply in your case, because you are the worst kind of crank - the "aggressively ignorant" kind, whose narcissistic "I know I am right and have discovered this revolutionary thing which all scientists in history have missed" conceit is exceeded only by his utter cluelessness.

The fact that you have no clue about even elementary physics/math concepts like displacement and have shown no interest in attempting the physics you claim to "debunk" paints you as a hopeless case. All your attempts at "clever ripostes" are negated by your loud and repeated demonstrations of willful ignorance.

I mean, you couldn't even be bothered to peruse the Wikipedia entry on the Twin Paradox before posting your latest inanity-thread. Pathetic.

[kladner]

Thanks for mentioning the Wiki article for which I saw a link, but did not follow up. I am quite willing to state that I did not know these things as clearly as I thought. It is also clear that Heinlein was riffing off the Langevin example.

The traveler remains in a projectile for one year of his time, and then reverses direction. Upon return, the traveler will find that he has aged two years, while 200 years have passed on Earth. During the trip, both the traveler and Earth keep sending signals to each other at a constant rate, which places Langevin's story among the Doppler shift versions of the twin paradox. The relativistic effects upon the signal rates are used to account for the different aging rates. The asymmetry that occurred because only the traveler underwent acceleration, is used to explain why there is any difference at all, because "any change of velocity, or any acceleration has an absolute meaning".[A 6]

Also, in reference to my previous statement of ignorance-

Consider a space ship traveling from Earth to the nearest star system outside of our solar system: a distance d = 4 light years away, at a speed v = 0.8c (i.e., 80 percent of the speed of light).
(To make the numbers easy, the ship is assumed to attain its full speed immediately upon departure—actually it would take close to a year accelerating at 1 g to get up to speed.)

Full article here:
http://en.wikipedia.org/wiki/Twin_paradox

[Mr. P-1]

The traditional twin paradox is easily resolved by noting that there is no single inertial frame of reference with respect to which which the travelling twin is stationary throughout his journey. In fact, with respect to any IFoR, the travelling twin will be travelling faster than the stay-at-home twin for at least one of the legs of his journey. Moreover this leg will dominate the calculation.

However, consider the following variation: In a closed universe, the travelling twin circumnavigates it without ever accelerating. Of course, from his inertial point of view it is the stay-at-home twin who has circumnavigated the universe. So what do their respective clocks show when ultimately they meet again?

[davar55]

Originally Posted by ewmayer

"If I take a long drive in my car and end up at the same place I started, my odometer should show no change, right?"

Dave, any instinct I might have had to "be patient with the physics neophyte" does not apply in your case, because you are the worst kind of crank - the "aggressively ignorant" kind, whose narcissistic "I know I am right and have discovered this revolutionary thing which all scientists in history have missed" conceit is exceeded only by his utter cluelessness.

The fact that you have no clue about even elementary physics/math concepts like displacement and have shown no interest in attempting the physics you claim to "debunk" paints you as a hopeless case. All your attempts at "clever ripostes" are negated by your loud and repeated demonstrations of willful ignorance.

I mean, you couldn't even be bothered to peruse the Wikipedia entry on the Twin Paradox before posting your latest inanity-thread. Pathetic.

I had already read it.

The odometer measures distance traveled, a scalar. Non-zero of course.
It does not measure displacement, a vector from the spatial origin
of your frame of reference to your location at each instance of time,
hence of magnitude zero after a round trip.

I suggest you are simply wrong. Without the usual sarcasm, you're
welcome to challenge my "solution' of the paradox. But displacement
is indeed the key to displacing your concept of the twin's puzzle.

I strongly believe your second and third paragraphs above are for
their writer a sell-portrait.

You challenge harshly with hardly a clue.

[davar55]

This leads to my point:

Consider a space ship traveling from Earth to the nearest star system outside of our solar system: a distance d = 4 light years away, at a speed v = 0.8c (i.e., 80 percent of the speed of light).
(To make the numbers easy, the ship is assumed to attain its full speed immediately upon departure—actually it would take close to a year accelerating at 1 g to get up to speed.)

My point is that ignoring acceleration going or coming back
invalidates the analysis totally.

[ewmayer]

Originally Posted by davar55

I had already read it.

...and comprehended it ... not.

The odometer measures distance traveled, a scalar.

"Total displacement" along a path traveled is the clue - just as one derives scalar speed from vector velocity, one can get differential (scalar) distance from differential (vector) displacement, and integrate that to get "total displacement", which to anyone possessing a clue obviously equates to total distance traveled.

Which figures more prominently in special (and general, but no need for the full-blown theory here) relativity - speed or velocity?

[davar55]

Originally Posted by ewmayer

...and comprehended it ... not.

"Total displacement" along a path traveled is the clue - just as one derives scalar speed from vector velocity, one can get differential (scalar) distance from differential (vector) displacement, and integrate that to get "total displacement", which to anyone possessing a clue obviously equates to total distance traveled.

Which figures more prominently in special (and general, but no need for the full-blown theory here) relativity - speed or velocity?

No, total displacement is independent of the path, and only depends on
the starting point and ending point. Any other answer is simply
missing the point of the trick question.

And the instantaneous vector velocity is defined as an instantaneous
speed in an instantaneous direction. While we can then re-derive the
speed from the velocity, the definitional dependence is the reverse -
first define straight-line scalar speed, and from this define vector
velocity.

One more time: suppose a trip goes from origin O = (0,0,0) to another
point A = (a,b,c), then to point B = (d,e,f) then back to O= (0,0,0).
The total displacement for the whole trip is a vector of magnitude zero.

If you integrate differential distance (scalar) over a path, you get
total distance (scalar) NOT total displacement (vector, which in
a round trip equals zero).

Velocity is used in the Lorentz derivation, but the three dimensions
of vector v are scalars v_x, v_y, v_z.
So they're equally important.

And repeatedly claiming my cluelessness is beneath you and
this forum we both enjoy.

[chalsall]

Originally Posted by davar55

Any other answer is simply missing the point of the trick question.

May I respectfully suggest that you might not be the right person to be asking trick questions?

A good friend of mine once advised: "Be responsible to the listening into which you are speaking.